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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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$f(x)= \large\frac{e^x}{1+e^x}$ $\;I_1= \int \limits_{f(-a)}^{f(a)} x g[(x) (1-x)].dx;\;I_2= \int \limits_{f(-a)}^{f(a)} x g[(x) (1-x)].dx;$ then $\large\frac{I_1}{I_2}=?$

\[\begin {array} {1 1} (a)\;\frac{1}{4} \\ (b)\;\frac{1}{3} \\ (c)\;\frac{1}{2} \\ (d)\;2 \end {array}\]

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1 Answer

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By using property,
$f(x)=f(a+b-x)$
$a+b-x= f(a)+f(-a)-x$
$\qquad= \large\frac{e^a}{1+e^a}+\frac{e^{-a}}{1+e^{-a}}$$-x$
$\qquad= \large\frac{e^a}{1+e^a}+\frac{1}{e^{a}}\frac{e^{-a}}{1+e^{-a}}$$-x$
$a+b-x=(1-x)$
$\;I_1= \int \limits_{f(-a)}^{f(a)} x g[(x) (1-x)].dx$
$\;I_1= \int \limits_{f(-a)}^{f(a)} (1-x) g[(1-x)x].dx$
$I_1=I_2-I_1$
$\large\frac{I_1}{I_2}= \frac{1}{21}$
answered Dec 13, 2013 by meena.p
 

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